If y(x) is a solution of the differential equation (2+sinx/1+y)dy/dx = - cos x and y(0) = 1, then find the value of y

Question

If y(x) is a solution of the differential equation \((\frac{2+sinx}{1+y})\) \(\frac{dy}{dx}\) = - cos x and y(0) = 1, then find the value of y \((\frac{\pi}{2}).\)

Answer 1

Given, differential equation is

⇒ log |1 + y| + log|2 + sin x| = log C

⇒ log (|1 + y| |2 + sin x|) = log C [∵ log m + log n = log mn]

⇒ (1 + y) (2 + sin x) = C .............. (i)

Also, given that at x = 0, y = 1.

On putting x = 0 and y = 1 in Eq. (i), we get

(1 + 1) (2 + sin 0) = C ⇒ C = 4

On putting C = 4 in Eq. (i), we get

(1 + y) (2 + sin x) = 4